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the resistance increasing gradually from o to B c. But the force
must be multiplied by the arm to give the moment. The lamina
A B c D represents the moment for the outer ring, being

/2  IL P \

force ( -£— j  x arm (r)

Similarly ab cd\s> the moment at ring r2, and the pyramid volume
will give the total moment, thus :

2 u. P          r       2

Moment of friction =         x *• x - = - aPr

-----------------------        r           3      3r

If P and r are in Ibs. and inches, the moment is in pound inches,
and the distribution of pressure may be such as to reduce it to \
fjL Pr. Concentrating the total force at the outer ring, 'it will be

"^ and     Work lost per m.   =   f M P x 2 TrRN
which may decrease to £ /i P x 2 ?r R N

Example 55.—Find H.P. lost in a footstep, whose dia. is 4", total
load 3000 Ibs., revs. 100 per m., when /* = "06. (Hons. Mach. Constr.
Ex., 1887.)

H. P. lost =

2 X "06 X 3OOO X 2 X 22 X 2 X IOO

3 x 33000 x 7 x 12

Example 56. — Mean dia. of thrust bearing = 14", screw thrust
40,000 Ibs., and pitch 15 ft. /* = '003, and 1000 miles are travelled in
3^ days. Find H. P. lost in friction. (Eng. Ex., 1888.)

Speed of vessel

and as vessel travels 15 ft. per rev.


ft. per m.

Revs, per m. =

3*5 x 24x60 x 15

.-. HvP.lost=^--^^--«
-----------        33000


33000 x 7 x 12 x 3'5 x 24 x 60 x 15

i- The form of pivot surface may be flat, conical, spherical, or

specially formed.     If conical, -.— must be substituted for P,

where a = angle at cone apex.

(Seep. 507.)